Let $G$ be a connected, undirected graph. A cut in $G$ is a set of edges whose removal results in $G$ being broken into two or more components, which are not connected with each other. The size of a cut is called its cardinality. A min-cut of $G$ is a cut in $G$ of minimum cardinality. Consider the following graph.

Which of the following sets of edges is a cut

$\left\{(A, B), (E, F), (B, D), (A, E), (A, D)\right\}$

$\left\{(B, D), (C, F), (A, B)\right\}$

What is cardinality of min-cut in this graph?

Prove that if a connected undirected graph $G$ with $n$ vertices has a min-cut of cardinality $k$, then $G$ has at least $\left(\frac{n\times k}{2}\right)$ edges.

5 Answers

Not a cut. We have spanning tree after removing this edges.
This is cut.we break graph into two pieces.

Min cut size is $2. \{BC,CF\}$. Removal these two edges disconnects $C$ from the remaining graph.

Always cardinality of min-cut will be $\leq$ min-degree of the graph
So, $k \leq$ min-degree of the graph
min-degree of the graph $\geq k$
WKT sum of degrees of all vertices $= 2 *$ number of edges.
minimum degree $\times n \leq 2\times |E|$, where $n$ is number of vertices, $|E|$ is number of edges.
$=k*n \leq 2|E|$

2- it will disconnect the graph. and remember the disconnected graph can have more than one vertices

answer for B,

we can always make a graph disconnected by removing the edges equal to min degree of a node. not believe me . remove the two edges BC and CF . as we can get a islolated vertex. tit will become disconnected .

on the same basis the proof can be done,

if k edges has to be removed , to make a graph disconnected . that means . edge connectivity should be ;less than equal to min degree edges.